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be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
+1 w
, such that 
and 
.
such that 
is prime, 
with 
and ?
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
can be placed tangent to side 
of 
such that the first circle is tangent to 
, subsequent circles are externally tangent to each other, and the last is tangent to 
. Similarly, 
circles of radius 
can also be placed along in this manner. The inradius of is 
, where 
and are relatively prime positive integers. Find 
.
, where 
. Each of the 
subsets of 
is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set 
, we then write 
for the number of subsets of 
that are blue.
and 
of , ![\[ f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2). \]](http://latex.artofproblemsolving.com/1/8/1/18184e9196e6c73601653d85f1cfd6d9798a8e43.png)
, let 
and 
be points on the side such that 
. If 
and 
are, respectively, the points of tangency of the incircles of the triangles 
and 
with the line , then show that![\[\frac{1}{MB}+\frac{1}{MD}= \frac{1}{NC}+\frac{1}{NE}. \]](http://latex.artofproblemsolving.com/0/b/b/0bbb21885659585aa235e2f5980099bbb533419f.png)
year revive, but hey, this solution is different!![\[
\frac{1}{s-b}+\frac{1}{s-c} = \frac{a}{(s-b)(s-c)} = \frac{as(s-a)}{K^2} = \frac{as(s-a)}{\frac12ah_a\cdot \frac12rs} = \frac{4(s-a)}{h_ar} = \frac{4\cot\frac12 A}{h_a},
\]](http://latex.artofproblemsolving.com/b/4/b/b4bfe230e07b2bb5db7e72796cffae378230ffab.png)
where 
is the height of the altitude from 
. Since and are the same for both triangle and triangle , the result follows.
,
,
denoting 
as the circunradium of triangle 
,
,
but by the Law of Sines we know that 
consequently,
therefore,
,
considering that 
we get,
yields 
consequently,
considering that 
we get,
and,
and the altitude 
.![\[ {1\over MB}+{1\over MD}={BD\over MB\cdot MD}={4BD\over (BD-AD+AB)(BD+AD-AB)}={4BD\over BD^2-(AD-AB)^2}={2BD\over AB\cdot AD(1-\cos\alpha)}={BD \sin \alpha\over S(ABD)(1-\cos \alpha)}={2\sin \alpha\over h(1-\cos\alpha)}={1\over NC}+{1\over NE} \]](http://latex.artofproblemsolving.com/e/c/5/ec5523089304fba022522d185e67c00d2677838b.png)
,
,
,
are 
,
,
,
,
denothe the area, then: 
:
![$ S = \frac {mn}{2} sin\frac {\alpha}{2} + \frac {1}{2}[rctg\frac {\alpha}{2}(m + n + rctg\frac {\alpha}{2})]sin\alpha$](http://latex.artofproblemsolving.com/9/1/e/91ed31f5fd7c5b1774cab340353cea7db70edc0c.png)
; 
![$ \frac {[ABD]}{BD} = \frac {[ACE]}{CE}$](http://latex.artofproblemsolving.com/6/e/1/6e16c4be1c4a48b1f7d846e3b066f5d3e81d9505.png)
